Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt = k*v^2.?
Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt = k*v^2. The initial speed of the motorboat is v(0) = 10 m/s, and v is decreasing at the rate of 1 m/s^2 when v=5 m/s. (a) How long does it take for the velocity of the boat to decrease to 1 m/s? (b) To (1/10) m/s? (c) When does the boat come to a stop?
- hfshawLv 71 decade agoBest Answer
You have that:
dv/dt = k*v^2
and you should already be able to tell that k is going to be a negative number.
You are told that when v = 5 m/s, dv/dt = -1 m/s^2, so:
-1 m/s^2 = k*(5 m/s)^2
k = -1/(25 m)
The original differential equation is separable:
v^(-2) dv = k dt
which integrates to:
-1/v = k*t + c
where c is the constant of integration.
v(t) = -1/(k*t+c)
You are told that at t = 0, v(t) = 10m/s, so:
10 m/s = -1/c
c = -0.1 s/m
Plugging in the values for k and c, we have that the equation for the velocity is:
v(t) = 1/(0.1 s/m + t/25m)
The velocity of the boat is 1 m/s when:
1m/s = 1/(0.1 s/m + t/25m)
1s/m = (0.1 s/m + t/25m)
25 s = 2.5s + t
t = 22.5 sec
The velocity of the boat is 0.1 m/s when:
0.1 m/s = 1/(0.1 s/m + t/25m)
10 s/m = 0.1 s/m + t/25m
250s = 2.5 s + t
t = 247.5 sec
The boat's speed is zero only in the limit as t -> infinity.